MOMENTS OF LOGCOS AND LOGSINE

Some fun integrals for Friday:

$\int_0^{\pi/2}x \ln\left(\cos(x)\right)\,dx=-\frac{\pi^2\ln(2)}{8}-\frac{7\zeta(3)}{16}$

$\int_0^{\pi/2}x \ln\left(\sin(x)\right)\,dx=-\frac{\pi^2\ln(2)}{8}+\frac{7\zeta(3)}{16}$

$\begin{aligned} I&=\int_0^{\pi/2}x \ln\left(\cos(x)\right)\,dx\\ &=-\ln(2)\int_0^{\pi/2}x \,dx+\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \int_0^{\pi/2}x \cos(2 k x)\,dx\\ &=-\frac{\pi^2\ln(2)}{8}+\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \left(\frac{x \sin(2 k x)}{2k}\Big|_0^{\pi/2} -\frac{1}{2k}\int_0^{\pi/2}\sin(2 k x)\,dx\right)\\ &=-\frac{\pi^2\ln(2)}{8}+\frac{1}{2}\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2} \left(\frac{\cos(2 k x)}{2k}\Big|_0^{\pi/2}\right)\\ &=-\frac{\pi^2\ln(2)}{8}+\frac{1}{4}\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3} \left((-1)^k-1\right)\\ &=-\frac{\pi^2\ln(2)}{8}-\frac{1}{4}\zeta(3)-\frac14\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3}\\ &=-\frac{\pi^2\ln(2)}{8}-\frac{\zeta(3)}{4}-\frac{\eta(3)}{4}\\ &=-\frac{\pi^2\ln(2)}{8}-\frac{\zeta(3)}{4}-\frac{3\zeta(3)}{16}\\ &=-\frac{\pi^2\ln(2)}{8}-\frac{7\zeta(3)}{16} \qquad \blacksquare\\ \end{aligned}$

$\begin{aligned} I&=\int_0^{\pi/2}x \ln\left(\sin(x)\right)\,dx\\ &=-\ln(2)\int_0^{\pi/2}x \,dx-\sum_{k=1}^\infty \frac{1}{k} \int_0^{\pi/2}x \cos(2 k x)\,dx\\ &=-\frac{\pi^2\ln(2)}{8}-\sum_{k=1}^\infty \frac{1}{k} \left(\frac{x \sin(2 k x)}{2k}\Big|_0^{\pi/2} -\frac{1}{2k}\int_0^{\pi/2}\sin(2 k x)\,dx\right)\\ &=-\frac{\pi^2\ln(2)}{8}-\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k^2} \left(\frac{\cos(2 k x)}{2k}\Big|_0^{\pi/2}\right)\\ &=-\frac{\pi^2\ln(2)}{8}-\frac{1}{4}\sum_{k=1}^\infty \frac{1}{k^3} \left((-1)^k-1\right)\\ &=-\frac{\pi^2\ln(2)}{8}+\frac{1}{4}\zeta(3)+\frac14\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3}\\ &=-\frac{\pi^2\ln(2)}{8}+\frac{\zeta(3)}{4}+\frac{\eta(3)}{4}\\ &=-\frac{\pi^2\ln(2)}{8}+\frac{\zeta(3)}{4}+\frac{3\zeta(3)}{16}\\ &=-\frac{\pi^2\ln(2)}{8}+\frac{7\zeta(3)}{16} \qquad \blacksquare\\ \end{aligned}$